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MHT CET 2021 (23 Sep, Shift 2) · Current Loop as a Magnetic Dipole

Current Loop as a Magnetic Dipole question from MHT CET 2021 (23 Sep, Shift 2)

Q50MHT CET 2021 · 23 SEP · SHIFT 2Easy

Current Loop as a Magnetic Dipole

An electron in a circular orbit of radius 0.05 nm0.05\text{ nm} performs 101610^{16} revolutions/second. What is the magnetic moment due to the rotation of electron? (e=1.6×10−19 C)(e = 1.6 \times 10^{-19}\text{ C})

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Answer: (d)

  1. I=ef=1.6×10−19×1016=1.6×10−3 AI = ef = 1.6\times10^{-19}\times10^{16} = 1.6\times10^{-3}\text{ A}
  2. A=πr2=π(5×10−11)2=7.85×10−21 m2A = \pi r^2 = \pi(5\times10^{-11})^2 = 7.85\times10^{-21}\text{ m}^2
  3. μ=IA=1.6×10−3×7.85×10−21=1.26×10−23 A m2\mu = IA = 1.6\times10^{-3}\times7.85\times10^{-21} = 1.26\times10^{-23}\text{ A m}^2
  4. So the answer is (d).
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